Understanding Negative Current in Discharging Capacitors

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

A)Clearly,
$V_A - V_B >0 \\V_C - V_D >0$ as the current flows from higher potential to the lower potential.
As the time passes on, the potential difference across the capacitor as well as the resistor goes on decreasing. Consequently, according to Ohm's law, the current goes on decreasing. This means that either the speed of the charged particles is decreasing or the no. of charged particles passing through the resistor is decreasing.
Assuming that it is the no. of charged particles which is decreasing, keeping the speed constant,
the current defined ,here , as I = $\frac {dq}{dt}$ is negative.

So, the concept "negative current" has two meanings : 1) $\frac {dq}{dt}$ is negative.
2) the current is flowing in the opposite direction. Right?

Now,
$[V_A - V_B= q/C ] = [ V_C - V_D = - \frac {dq}{dt} R$
Is this correct so far?

 

[cnh1995]

Is this correct so far?

Yes.

 

[cnh1995]

) dqdtdqdt \frac {dq}{dt} is negative.

This is correct.

the current is flowing in the opposite direction. Right?

No. In this case, the direction shown in the diagram is correct. So negative dq/dt means current reduces over time.